The greatest number for four digits which when divided by 12, 16 and 24 leave remainder 2,6 and 14 respectively is = ?
(A) 9974
(B) 9970
(C) 9807
(D) 9998
Solution:
12 - 2 = 10 16 - 6 = 10 24 - 14 = 10 LCM of (12, 16, 24) = 6 × 2 × 4 × 1 = 48 Greatest number of four digits = 9999 ∴ When it is divided by 48 we get remainder = 15 ⇒ The greatest number of 4 digits which completely divides the given number is = 9999 - 15 = 9984 ∴ Number is = 9984 - 10 = 9974
54.
The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is:
(A) 28
(B) 32
(C) 40
(D) 64
Solution:
Let the numbers be 2x and 3x Then, their L.C.M. = 6x So, 6x = 48 or x = 8 ∴ The numbers are 16 and 24 Hence, required sum = (16 + 24) = 40
55.
252 can be expressed as a product of primes as:
(A) 2 × 2 × 3 × 3 × 7
(B) 2 × 2 × 2 × 3 × 7
(C) 3 × 3 × 3 × 3 × 7
(D) 2 × 3 × 3 × 3 × 7
Solution:
Clearly, 252 = 2 × 2 × 3 × 3 × 7
56.
The sum of two numbers is 528 and their HCF is 33. The number of pairs of numbers satisfying the above conditions is = ?
(A) 4
(B) 6
(C) 8
(D) 12
Solution:
Let the required numbers be 33a and 33b. Then 33a + 33b = 528 ⇒ a + b = 16 Now co - primes with sum 16 are (1, 15), (3, 13), (5, 11) and (7, 9) ∴ Required numbers are (33 × 1, 33 × 15), (33 × 3, 33 × 13), (33 × 5, 33 × 11), (33 × 7, 33 × 9) The number of such pairs is 4.
57.
The HCF of $$\frac{9}{{10}}{\text{,}}\frac{{12}}{{25}}{\text{,}}\frac{{18}}{{35}}$$ ÃÂÃÂ and $$\frac{{21}}{{40}}$$ is = ?
(A) $$\frac{3}{5}$$
(B) $$\frac{{252}}{5}$$
(C) $$\frac{3}{{1400}}$$
(D) $$\frac{{63}}{{700}}$$
Solution:
$$\eqalign{ & {\text{Required HCF }} \cr & {\text{ = }}\frac{{{\text{HCF of 9,12,18,21}}}}{{{\text{LCM of 10,25,35,40}}}} \cr & = \frac{3}{{1400}} \cr} $$
58.
Let x be the last number, which when divided by 5, 6, 7 and 8 leaves a remainder 3 in each case but when divided by 9 leaves remainder 0, the sum of digits of x is = ?
The least perfect square which is divisible by each of 21, 36 and 66 is = ?
(A) 214344
(B) 214434
(C) 213444
(D) 231444
Solution:
L.C.M. of (21, 36, 66) = 21 × 12 × 11 = 7 × 3 × 4 × 3 × 11 = 7 × 3 × 2 × 2 × 3 × 11 For perfect square multiply by 7 × 11 So that pairs of number from perfect square ∴ 7 × 7 × 3 × 3 × 2 × 2 × 11 × 11 Required result is = 213444 (which is a perfect square)
60.
The LCM of two prime numbers x and y (x > y) is 161. The value of (3y - x):
(A) -2
(B) -1
(C) 1
(D) 2
Solution:
LCM of prime number x and y = 161 (given) ⇒ xy = 161 ⇒ xy = 23 × 7 ⇒ x = 23, y = 7 (∵ x > y given) So, the value of (3y - x) = 3 × 7 - 23 = 21 - 23 = -2